How would I find the parametric equations for the line through $𝑃_0=(9,−1,1)$ perpendicular to the plane $10𝑥+12𝑦−4𝑧=10.$?
Given:
$𝑥=9+10𝑡$
What is $y$ and $z$?
So first what I do is find the normal vector to the plane which is as follows:
$\left\langle10,12,-4\right\rangle$
Then I used that to find the parametrization of the line: $𝑃_0=(9,−1,1)$
and so I get $y=-1+12t$ and $z=1+-4t$.right or wrong
Best Answer
The direction vector for this line is $v=(10,12,-4)$ (i.e normal vector is $N=10i+12j-4k$) and it must pass through the point $P_{0}=(9,-1,1).$
Thus we have parametric equations $$(x,y,z)=(9,-1,1)+(10,12,-4)t=(9+10t,-1+12t,1-4t).$$